Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T sqr(T a) {
  return a * a;
};
template <typename T>
void dprint(T begin, T end) {
  for (auto i = begin; i != end; i++) cerr << (*i) << " ";
  cerr << "\n";
};
const int maxn = 2e5 + 10;
const double eps = 1e-8;
int sgn(double x) { return x < -eps ? -1 : x > eps ? 1 : 0; }
struct Point {
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  void print() { printf("(%.2lf, %.2lf)\n", x, y); }
  Point operator+(Point p) { return Point(x + p.x, y + p.y); }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  Point operator*(double d) { return Point(x * d, y * d); }
  bool operator<(Point p) const { return x != p.x ? x < p.x : y < p.y; }
  double dot(Point p) { return x * p.x + y * p.y; }
  double det(Point p) { return x * p.y - y * p.x; }
  Point ver() { return Point(-y, x); }
};
Point intersection(Point p1, Point p2, Point q1, Point q2) {
  return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}
int n, m;
double l[maxn], r[maxn], sum[maxn], sy, a, b;
Point A, B;
int main() {
  scanf("%lf%lf%lf", &sy, &a, &b);
  A = Point(a, sy);
  B = Point(b, sy);
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    scanf("%lf%lf", l + i, r + i), sum[i + 1] += sum[i] + r[i] - l[i];
  scanf("%d", &m);
  for (int i = 0, x, y; i < m; i++) {
    scanf("%d%d", &x, &y);
    Point p = Point(x, y);
    Point AA = intersection(p, A, Point(0, 0), Point(1, 0)),
          BB = intersection(p, B, Point(0, 0), Point(1, 0));
    int j = upper_bound(r, r + n, AA.x) - r;
    int k = upper_bound(r, r + n, BB.x) - r;
    double ans = 0;
    if (j < n) ans -= max(0., AA.x - l[j]);
    if (k < n) ans -= r[k] - max(l[k], BB.x);
    ans += sum[min(k + 1, n)] - sum[j];
    printf("%.12f\n", ans * (-sy + y) / y);
  }
  return 0;
}